Given four integers a, b, c representing coefficients of a straight line with equation (ax + by + c = 0), the task is to find the equations of the two straight lines passing through a given point 
and making an angle ? with the given straight line.
Examples:
Input: a = 2, b = 3, c = -7, x1 = 4, y1 = 9, ? = 30
Output: y = -0.49x +10
y = -15.51x + 71Input: a = 3, b = -2, c = 4, x1 = 3, y1 = 4, ? = 55
Output: y = 43.73x -127
y = -0.39x +5
Approach:
Figure 1
- Let P (x1, y1) be the given point and line LMN (In figure 1) be the given line making an angle ? with the positive x-axis.
- Let PMR and PNS be two required lines which makes an angle (?) with the given line.
- Let these lines meet the x-axis at R and S respectively.
- Suppose line PMR and PNS make angles (?1) and (?2) respectively with the positive direction of the x-axis.
- Then using the slope point form of a straight line, the equation of two lines are :
… (1)
and
… (1)
… (2)
and 
are the slopes of lines PMR and PNS respectively. 
Figure 2
- Now consider triangle LMR:
Using the property: An exterior angle of a triangle is equal to the sum of the two opposite interior angles
… (3)
… (3)- Now consider triangle LNS:
Figure 3
… (4)- Now we calculate the value of (tan?):
By formula,
slope of given line
slope of given line 
- Now substitute the values of (tan(?1)) and (tan(?2)) from equations (3) and (4) to equations (1) and (2) to get the final equations of both the lines:
Line PMR :
Line PNS :
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find slope of given linedouble line_slope(double a, double b){ if (a != 0) return -b / a; // Special case when slope of // line is infinity or is // perpendicular to x-axis else return (-2);}// Function to find equations of lines// passing through the given point// and making an angle with given linevoid line_equation(double a, double b, double c, double x1, double y1, double alfa){ // Set the precision cout << fixed << setprecision(2); // Store slope of given line double given_slope = line_slope(a, b); // Convert degrees to radians double x = alfa * 3.14159 / 180; // Special case when slope of // given line is infinity: // In this case slope of one line // will be equal to alfa // and the other line will be // equal to (180-alfa) if (given_slope == -2) { // In this case slope of // required lines can't be // infinity double slope_1 = tan(x); double slope_2 = tan(3.14159 - x); // g and f are the variables // of required equations int g = x1, f = x1; g *= (-slope_1); g += y1; // Print first line equation if (g > 0) cout << "y = " << slope_1 << "x +" << g << endl; if (g <= 0) cout << "y = " << slope_1 << "x " << g << endl; f *= (-slope_2); f += y1; // Print second line equation if (f > 0) { cout << "y = " << slope_2 << "x +" << f << endl; } if (f <= 0) cout << "y = " << slope_2 << "x " << f << endl; return; } // Special case when slope of // required line becomes infinity if (1 - tan(x) * given_slope == 0) { cout << "x = " << x1 << endl; } if (1 + tan(x) * given_slope == 0) { cout << "x = " << x1 << endl; } // General case double slope_1 = (given_slope + tan(x)) / (1 - tan(x) * given_slope); double slope_2 = (given_slope - tan(x)) / (1 + tan(x) * given_slope); // g and f are the variables // of required equations int g = x1, f = x1; g *= (-slope_1); g += y1; // Print first line equation if (g > 0 && 1 - tan(x) * given_slope != 0) cout << "y = " << slope_1 << "x +" << g << endl; if (g <= 0 && 1 - tan(x) * given_slope != 0) cout << "y = " << slope_1 << "x " << g << endl; f *= (-slope_2); f += y1; // Print second line equation if (f > 0 && 1 + tan(x) * given_slope != 0) { cout << "y = " << slope_2 << "x +" << f << endl; } if (f <= 0 && 1 + tan(x) * given_slope != 0) cout << "y = " << slope_2 << "x " << f << endl;}// Driver Codeint main(){ // Given Input double a = 2, b = 3, c = -7; double x1 = 4, y1 = 9; double alfa = 30; // Function Call line_equation(a, b, c, x1, y1, alfa); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{ // Function to find slope of given linestatic double line_slope(double a, double b){ if (a != 0) return -b / a; // Special case when slope of // line is infinity or is // perpendicular to x-axis else return (-2);}// Function to find equations of lines// passing through the given point// and making an angle with given linestatic void line_equation(double a, double b, double c, double x1, double y1, double alfa){ // Store slope of given line double given_slope = line_slope(a, b); // Convert degrees to radians double x = alfa * 3.14159 / 180; // Special case when slope of // given line is infinity: // In this case slope of one line // will be equal to alfa // and the other line will be // equal to (180-alfa) if (given_slope == -2) { // In this case slope of // required lines can't be // infinity double slope_1 = Math.tan(x); double slope_2 = Math.tan(3.14159 - x); // g and f are the variables // of required equations int g = (int)x1, f = (int)x1; g *= (-slope_1); g += y1; // Print first line equation if (g > 0) System.out.println("y = " + (Math.round(slope_1 * 100.0) / 100.0) + "x +" + (Math.round(g * 100.0) / 100.0)); if (g <= 0) System.out.println("y = " + (Math.round(slope_1 * 100.0) / 100.0) + "x " + (Math.round(g * 100.0) / 100.0)); f *= (-slope_2); f += y1; // Print second line equation if (f > 0) { System.out.println("y = " + (Math.round(slope_2 * 100.0) / 100.0) + "x +" + (Math.round(f * 100.0) / 100.0)); } if (f <= 0) System.out.println("y = " + (Math.round(slope_1 * 100.0) / 100.0) + "x " + (Math.round(g * 100.0) / 100.0)); return; } // Special case when slope of // required line becomes infinity if (1 - Math.tan(x) * given_slope == 0) { System.out.println("x = " + (Math.round(x1 * 100.0) / 100.0)); } if (1 + Math.tan(x) * given_slope == 0) { System.out.println("x = " + (Math.round(x1 * 100.0) / 100.0)); } // General case double slope_1 = (given_slope + Math.tan(x)) / (1 - Math.tan(x) * given_slope); double slope_2 = (given_slope - Math.tan(x)) / (1 + Math.tan(x) * given_slope); // g and f are the variables // of required equations int g = (int)x1, f = (int)x1; g *= (-slope_1); g += y1; // Print first line equation if (g > 0 && 1 - Math.tan(x) * given_slope != 0) System.out.println("y = " + (Math.round(slope_1 * 100.0) / 100.0) + "x +" + (Math.round(g * 100.0) / 100.0)); if (g <= 0 && 1 - Math.tan(x) * given_slope != 0) System.out.println("y = " + (Math.round(slope_1 * 100.0) / 100.0) + "x " + (Math.round(g * 100.0) / 100.0)); f *= (-slope_2); f += y1; // Print second line equation if (f > 0 && 1 + Math.tan(x) * given_slope != 0) { System.out.println("y = " + (Math.round(slope_2 * 100.0) / 100.0) + "x +" + (Math.round(f * 100.0) / 100.0)); } if (f <= 0 && 1 + Math.tan(x) * given_slope != 0) System.out.println("y = " + (Math.round(slope_2 * 100.0) / 100.0) + "x +" + (Math.round(f * 100.0) / 100.0));}// Driver Codepublic static void main (String[] args) { // Given Input double a = 2, b = 3, c = -7; double x1 = 4, y1 = 9; double alfa = 30; // Function Call line_equation(a, b, c, x1, y1, alfa);}}// This code is contributed by Dharanendra L V. |
Python3
# Python3 program for the above approachimport math# Function to find slope of given linedef line_slope(a, b): if (a != 0): return -b / a # Special case when slope of # line is infinity or is # perpendicular to x-axis else: return (-2)# Function to find equations of lines# passing through the given point# and making an angle with given linedef line_equation(a, b, c, x1, y1, alfa): # Store slope of given line given_slope = line_slope(a, b) # Convert degrees to radians x = alfa * 3.14159 / 180 # Special case when slope of # given line is infinity: # In this case slope of one line # will be equal to alfa # and the other line will be # equal to (180-alfa) if (given_slope == -2): # In this case slope of # required lines can't be # infinity slope_1 = math.tan(x) slope_2 = math.tan(3.14159 - x) # g and f are the variables # of required equations g = x1, f = x1 g *= (-slope_1) g += y1 # Print first line equation if (g > 0): print("y = ", round(slope_1, 2), "x +" , round(g)); if (g <= 0): print("y = ", round(slope_1, 2), "x ", round(g)) f *= (-slope_2) f += y1 # Print second line equation if (f > 0): print("y = ", round(slope_2, 2), "x +", round(f)) if (f <= 0): print("y = " , round(slope_2, 2), "x " , round(f)) return # Special case when slope of # required line becomes infinity if (1 - math.tan(x) * given_slope == 0): print("x =", x1) if (1 + math.tan(x) * given_slope == 0): print("x =", x1) # General case slope_1 = ((given_slope + math.tan(x)) / (1 - math.tan(x) * given_slope)) slope_2 = ((given_slope - math.tan(x)) / (1 + math.tan(x) * given_slope)) # g and f are the variables # of required equations g = x1 f = x1 g *= (-slope_1) g += y1 # Print first line equation if (g > 0 and 1 - math.tan(x) * given_slope != 0): print("y = ", round(slope_1, 2), "x +", round(g)) if (g <= 0 and 1 - math.tan(x) * given_slope != 0): print("y = ", round(slope_1, 2), "x ", round(g)) f *= (-slope_2) f += y1 # Print second line equation if (f > 0 and 1 + math.tan(x) * given_slope != 0): print("y = ", round(slope_2, 2), "x +", round(f)) if (f <= 0 and 1 + math.tan(x) * given_slope != 0): print("y = ", round(slope_2, 2), "x " , round(f)) # Driver Codeif __name__ == "__main__": # Given Input a = 2 b = 3 c = -7 x1 = 4 y1 = 9 alfa = 30 # Function Call line_equation(a, b, c, x1, y1, alfa)# This code is contributed by ukasp |
C#
// C# program for the above approachusing System;using System.Collections.Generic;public class GFG{ // Function to find slope of given linestatic double line_slope(double a, double b){ if (a != 0) return -b / a; // Special case when slope of // line is infinity or is // perpendicular to x-axis else return (-2);}// Function to find equations of lines// passing through the given point// and making an angle with given linestatic void line_equation(double a, double b, double c, double x1, double y1, double alfa){ // Store slope of given line double given_slope = line_slope(a, b); // Convert degrees to radians double x = alfa * 3.14159 / 180; double slope_1,slope_2; double g,f; // Special case when slope of // given line is infinity: // In this case slope of one line // will be equal to alfa // and the other line will be // equal to (180-alfa) if (given_slope == -2) { // In this case slope of // required lines can't be // infinity slope_1 = Math.Tan(x); slope_2 = Math.Tan(3.14159 - x); // g and f are the variables // of required equations g = (int)x1; f = (int)x1; g *= (-slope_1); g += y1; // Print first line equation if (g > 0) Console.WriteLine("y = " + (Math.Round(slope_1 * 100.0) / 100.0) + "x +" + (Math.Round((int)g * 100.0) / 100.0)); if (g <= 0) Console.WriteLine("y = " + (Math.Round(slope_1 * 100.0) / 100.0) + "x " + (Math.Round((int)g * 100.0) / 100.0)); f *= (-slope_2); f += y1; // Print second line equation if (f > 0) { Console.WriteLine("y = " + (Math.Round(slope_2 * 100.0) / 100.0) + "x +" + (Math.Round((int)f * 100.0) / 100.0)); } if (f <= 0) Console.WriteLine("y = " + (Math.Round(slope_1 * 100.0) / 100.0) + "x " + (Math.Round((int)g * 100.0) / 100.0)); return; } // Special case when slope of // required line becomes infinity if (1 - Math.Tan(x) * given_slope == 0) { Console.WriteLine("x = " + (Math.Round(x1 * 100.0) / 100.0)); } if (1 + Math.Tan(x) * given_slope == 0) { Console.WriteLine("x = " + (Math.Round(x1 * 100.0) / 100.0)); } // General case slope_1 = (given_slope + Math.Tan(x)) / (1 - Math.Tan(x) * given_slope); slope_2 = (given_slope - Math.Tan(x)) / (1 + Math.Tan(x) * given_slope); // g and f are the variables // of required equations g = (int)x1; f = (int)x1; g *= (-slope_1); g += y1; // Print first line equation if (g > 0 && 1 - Math.Tan(x) * given_slope != 0) Console.WriteLine("y = " + (Math.Round(slope_1 * 100.0) / 100.0) + "x +" + (Math.Round((int)g * 100.0) / 100.0)); if (g <= 0 && 1 - Math.Tan(x) * given_slope != 0) Console.WriteLine("y = " + (Math.Round(slope_1 * 100.0) / 100.0) + "x " + (Math.Round((int)g * 100.0) / 100.0)); f *= (-slope_2); f += y1; // Print second line equation if (f > 0 && 1 + Math.Tan(x) * given_slope != 0) { Console.WriteLine("y = " + (Math.Round(slope_2 * 100.0) / 100.0) + "x +" + (Math.Round((int)f * 100.0) / 100.0)); } if (f <= 0 && 1 + Math.Tan(x) * given_slope != 0) Console.WriteLine("y = " + (Math.Round(slope_2 * 100.0) / 100.0) + "x +" + (Math.Round((int)f * 100.0) / 100.0));}// Driver Codepublic static void Main(String[] args) { // Given Input double a = 2, b = 3, c = -7; double x1 = 4, y1 = 9; double alfa = 30; // Function Call line_equation(a, b, c, x1, y1, alfa);}}// This code contributed by shikhasingrajput |
Javascript
// JavaScript program for the above approach// Function to find slope of given linefunction line_slope(a, b){ if (a != 0) return -b / a; // Special case when slope of // line is infinity or is // perpendicular to x-axis else return (-2);}// Function to find equations of lines// passing through the given point// and making an angle with given linefunction line_equation(a, b, c, x1, y1, alfa){ // Store slope of given line let given_slope = line_slope(a, b); // Convert degrees to radians let x = alfa * 3.14159 / 180; // Special case when slope of // given line is infinity: // In this case slope of one line // will be equal to alfa // and the other line will be // equal to (180-alfa) if (given_slope == -2) { // In this case slope of // required lines can't be // infinity let slope_1 = Math.tan(x); let slope_2 = Math.tan(3.14159 - x); // g and f are the variables // of required equations let g = x1, f = x1; g = g*(-slope_1); g = g + y1; // Print first line equation if (g > 0) console.log("y = ", slope_1.toFixed(2), "x +", Math.floor(g)); if (g <= 0) console.log("y = ", slope_1.toFixed(2), "x ", Math.floor(g)); f = f*(-slope_2); f = f+y1; // Print second line equation if (f > 0) { console.log("y = ", slope_2.toFixed(2), "x +", Math.floor(f)); } if (f <= 0){ console.log("y = ", slope_2.toFixed(2), "x ", Math.floor(f)); } return; } // Special case when slope of // required line becomes infinity if (1 - Math.tan(x) * given_slope == 0) { console.log("x = ", x1.toFixed(2)); } if (1 + Math.tan(x) * given_slope == 0) { console.log("x = ", x1.toFixed(2)); } // General case let slope_1 = (given_slope + Math.tan(x)) / (1 - Math.tan(x) * given_slope); let slope_2 = (given_slope - Math.tan(x)) / (1 + Math.tan(x) * given_slope); // g and f are the variables // of required equations let g = x1, f = x1; g *= (-slope_1); g += y1; // Print first line equation if (g > 0 && 1 - Math.tan(x) * given_slope != 0) console.log("y = ", slope_1.toFixed(2), "x +", Math.floor(g)); if (g <= 0 && 1 - tan(x) * given_slope != 0) console.log("y = ", slope_1.toFixed(2), "x ", Math.floor(g)); f *= (-slope_2); f += y1; // Print second line equation if (f > 0 && 1 + Math.tan(x) * given_slope != 0) { console.log("y = ", slope_2.toFixed(2), "x +", Math.floor(f)); } if (f <= 0 && 1 + tan(x) * given_slope != 0) console.log("y = ", slope_2.toFixed(2), "x ", Math.floor(f));}// Driver Code// Given Inputlet a = 2, b = 3, c = -7;let x1 = 4, y1 = 9;let alfa = 30;// Function Callline_equation(a, b, c, x1, y1, alfa);// The code is contributed by Gautam goel (gautamgoel962) |
Output:
y = -0.49x +10 y = -15.51x +71
Time Complexity: O(1)
Auxiliary Space: O(1)
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